Hyperbolas
By the end of this lesson, you’ll be able to:
- Identify the key features of a hyperbola (center, vertices, asymptotes, transverse axis, foci).
- Write and interpret hyperbolas in standard form.
- Determine whether a hyperbola opens horizontally or vertically.
- Connect algebraic parameters to geometric features, a skill commonly tested on ACT Math.
Key Ideas
A hyperbola is the set of all points for which the difference of the distances to two fixed points (the foci) is constant.
The standard forms:
Horizontal transverse axis: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]
Vertical transverse axis: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]
Where:
- \((h,k)\) is the center
- \(a\) determines distance from center to vertices
- \(b\) helps determine asymptotes
- Relationship: \[ c^2 = a^2 + b^2 \]
- Foci:
- Horizontal: \((h \pm c,\, k)\)
- Vertical: \((h,\, k \pm c)\)
- Horizontal: \((h \pm c,\, k)\)
Asymptotes:
Horizontal hyperbola: \[ y-k = \pm \frac{b}{a}(x-h) \]
Vertical hyperbola: \[ y-k = \pm \frac{a}{b}(x-h) \]
Common Problem Types
1. Identifying Orientation
Check which variable term is positive.
- Positive \(x\)-term → horizontal opening
- Positive \(y\)-term → vertical opening
2. Finding \(a\), \(b\), and \(c\)
Use \[ c^2 = a^2 + b^2 \] then determine the foci.
3. Finding Asymptotes
Use the asymptote formulas and center coordinates.
4. Converting from General Form
Complete the square for both variables and rewrite in standard form.
5. Writing Equations from Features
ACT problems may give vertices, foci, or asymptotes and ask for the equation.
Strategies
- Look for subtraction. Hyperbolas always involve a minus sign between squared terms.
- Orientation shortcut: The positive term determines opening direction.
- Do not confuse ellipses and hyperbolas.
- Ellipse: \[ c^2 = a^2 - b^2 \]
- Hyperbola: \[ c^2 = a^2 + b^2 \]
- Sketch asymptotes lightly before graphing branches.
- Check center signs carefully after completing the square.
Worked Examples
Example 1: Identify key features
Find the center, orientation, and foci of: \[ \frac{(x-1)^2}{16} - \frac{(y+2)^2}{9} = 1 \]
- Positive term is the \(x\)-term → horizontal hyperbola
- Center: \((1,-2)\)
- \(a^2 = 16 \Rightarrow a = 4\)
- \(b^2 = 9 \Rightarrow b = 3\)
Compute \(c\): \[ \begin{split} c^2 &= a^2 + b^2 \\ &= 16 + 9 \\ &= 25 \end{split} \]
So: \[ c = 5 \]
Foci: \[ (1 \pm 5,\,-2) \]
Therefore, the foci are: \[ (-4,-2)\text{ and }(6,-2) \]
Example 2: Write the equation
A hyperbola has center \((0,0)\), vertical transverse axis, \(a=3\), and \(b=5\).
Since the hyperbola opens vertically, the positive term goes under the \(y\)-variable:
\[ \frac{y^2}{3^2} - \frac{x^2}{5^2} = 1 \]
So the equation is: \[ \frac{y^2}{9} - \frac{x^2}{25} = 1 \]
Example 3: Convert to standard form
Convert \[ 4x^2 - 9y^2 - 16x - 54y - 101 = 0 \] to standard form.
Group terms: \[ 4(x^2 - 4x) - 9(y^2 + 6y) = 101 \]
Complete the square: \[ \begin{split} x^2 - 4x &= (x-2)^2 - 4 \\ y^2 + 6y &= (y+3)^2 - 9 \end{split} \]
Substitute: \[ 4[(x-2)^2 - 4] - 9[(y+3)^2 - 9] = 101 \]
Expand: \[ 4(x-2)^2 - 16 - 9(y+3)^2 + 81 = 101 \]
Combine constants: \[ 4(x-2)^2 - 9(y+3)^2 = 36 \]
Divide by 36: \[ \frac{(x-2)^2}{9} - \frac{(y+3)^2}{4} = 1 \]
This hyperbola has center \((2,-3)\) and opens horizontally.
- Forgetting that hyperbolas involve subtraction
- Using the ellipse formula \(c^2 = a^2 - b^2\) instead of \[ c^2 = a^2 + b^2 \]
- Mixing up horizontal and vertical forms
- Forgetting asymptotes when graphing
- Losing negative signs while completing the square
Practice Problems
Find the foci of \[ \frac{(x+2)^2}{25} - \frac{(y-1)^2}{16} = 1 \]
Write the equation of a hyperbola with center \((3,-4)\), vertical transverse axis, \(a=2\), \(b=6\).
Convert the hyperbola \[ 9x^2 - 4y^2 - 54x - 16y - 29 = 0 \] to standard form.
Identify whether the hyperbola opens horizontally or vertically: \[ \frac{(y+1)^2}{36} - \frac{(x-5)^2}{9} = 1 \]
A horizontal hyperbola has center \((0,0)\) and foci at \((\pm 13,0)\). If \(a=5\), find the equation.
1.
\(a^2=25\), \(b^2=16\)
\[ c^2 = 25 + 16 = 41 \]
\[ c=\sqrt{41} \]
Foci: \[ (-2 \pm \sqrt{41},\,1) \]
2.
Vertical transverse axis:
\[ \frac{(y+4)^2}{2^2} - \frac{(x-3)^2}{6^2} = 1 \]
Equation: \[ \frac{(y+4)^2}{4} - \frac{(x-3)^2}{36} = 1 \]
3.
Standard form: \[
\frac{(x-3)^2}{4} - \frac{(y+2)^2}{9} = 1
\]
Center: \[ (3,-2) \]
4.
Positive term is the \(y\)-term → vertical hyperbola.
5.
Foci at \((\pm 13,0)\): \[
c=13
\]
Given: \[ a=5 \Rightarrow a^2=25 \]
Use: \[ c^2 = a^2 + b^2 \]
\[ 169 = 25 + b^2 \]
\[ b^2 = 144 \]
Equation: \[ \frac{x^2}{25} - \frac{y^2}{144} = 1 \]
Summary
- Hyperbolas always contain a subtraction sign between squared terms.
- The positive term determines opening direction.
- Use \[
c^2 = a^2 + b^2
\] to find the foci.
- Asymptotes are a defining feature of hyperbolas.
- ACT problems often test orientation, graph interpretation, asymptotes, and converting from general form.
- Positive \(x\)-term → horizontal opening
- Positive \(y\)-term → vertical opening
- Hyperbola = subtraction
- Ellipse = addition
- Sketch asymptotes first when graphing